Trivial answer: take a finite subset. You are looking for any infinite subset?
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The UserMay 26 '13 at 18:18

@Benjamin It is not the case for every choosen subset $T\subset S$. For example consider $S=\left\{0\right\}\cup\left\{\frac{1}{n}\mid n\in\mathbb{N}\right\}$. If you remove the $0$, this subset is not order isomorphic to $S$.
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The UserMay 26 '13 at 18:30

The answer is yes in ZFC. We can construct a dense infinite set
$A\subset\mathbb{R}$ such that the only order-preserving map
$f:A\to A$ is the identity. In particular, $A$ is not
order-isomorphic with any proper subset of itself.

To see this, note first that any order-preserving map
$f:B\to\mathbb{R}$ defined on a dense set $B\subset\mathbb{R}$ can
be extended to a total order-preserving map $\bar
f:\mathbb{R}\to\mathbb{R}$ defined on the closure of $B$, by
defining $\bar f(x)=\sup_{y\leq x, y\in B}f(y)$. Further, note
that any such monotone map will have at most countably many points
of discontinuity, since every discontinuity will be a jump
discontinuity. Thus, there are precisely continuum many such
order-preserving functions $\mathbb{R}\to\mathbb{R}$, since any
one of them is determined by countably much information about
their values on a countable dense set and the information about
what their values are on the countably many points of
discontinuity.

We may therefore enumerate all order-preserving functions
$f_\alpha:\mathbb{R}\to\mathbb{R}$ in a sequence of length
continuum, $\alpha\lt\mathfrak{c}$.

Let's now build the set $A$ by a transfinite process, making
promises at each stage about some reals being definitely in $A$
and other promises about keeping some reals out of $A$, in such a
way that we kill off $f_\alpha$ at stage $\alpha$ as a possible
order-preserving map from $A$ to $A$. We may begin at stage $0$ by
placing all the rational numbers into $A$, so that it will
definitely be dense. Suppose we have carried out our process up to
stage $\alpha$, and $f_\alpha$ is the next non-identity
order-preserving map $\mathbb{R}\to\mathbb{R}$ presented for our
consideration. Since $f_\alpha$ is order-preserving and not the
identity, it must be that there is an interval $(a,b)$ with
$(f(a),f(b))$ disjoint from $(a,b)$. Since we've made fewer than
continuum many promises so far, there must be an $x\in (a,b)$ such
that we've made no promises about $x$ or $f_\alpha(x)$. In this
case, we place $x$ into $A$ and promise to keep $f_\alpha(x)$ out
of $A$. This will prevent $f_\alpha$ from being an
order-isomorphism of $A$ to a proper subset of $A$.

The end result is that $A$ is dense, but is strongly rigid in the
sense that there is no non-identity order-preserving map from $A$
to $A$. In particular, $A$ is not order-isomorphic with any proper
subset of itself.

EDIT: The other answers show that my intuition was wrong, and that in fact there is such a linear order in $ZFC$, so this answer (except the bit about determinacy) is superfluous.

Although it seems likely that $ZFC$ proves there is no such order, choice will certainly be necesary for such a proof: it is consistent with $ZF$ that $\mathbb{R}$ has infinite, Dedekind-finite subsets, which is exactly what you're asking for. (I'm looking for a reference . . . EDIT: Asaf gives a good reference in a comment, below.)

There can be no countable example of such an order, however; such an order can't embed $\mathbb{Q}$, and hence is scattered, and a result (I believe) of Jullien then lets us write it as a finite sum of indecomposable orders; it is then easy to see that the whole order embeds into a proper subset of itself.

If we assume the Axiom of Determinacy, then every uncountable set of reals has a perfect subset; this means there is no uncountable example of such an order, and hence by the above fact the answer to your question is no.

Under choice, things look a bit more complicated, but I suspect the answer is no; I'll post more if I can figure it out.

Thanks! I am interested in $\mathsf{ZFC}$, an explicit construction or at least an existence proof.
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Marty ColosMay 26 '13 at 18:38

Note that by work of Richard Laver, any such linear order has to not be a countable union of scattered orders. This is still massive overkill; I suspect there is a straightforward impossibility proof I'm not seeing.
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Noah SMay 26 '13 at 18:40

1

Also, note that (assuming large cardinals) we can't hope for an example which is "nicely definable" in the sense of being projective, for the same reason that $AD$ implies there is no such order at all. So this might mean, depending on your definition of "explicit," that there is no explicit construction.
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Noah SMay 26 '13 at 18:42

1

Noah, I like your answer very much, but your way of saying, "Yes, if we do not assume AC" could be misinterpreted as claiming that the answer is yes provably in ZF. But this is not what you mean, since you are claiming only the weaker assertion that a positive answer is relatively consistent with ZF.
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Joel David HamkinsMay 26 '13 at 20:23